# Question 78b35

Apr 25, 2017

$\text{Br"^(-) + 3"H"_3"AsO"_4 -> "BrO"_3^(-) + 3"HAsO"_2 + 3"H"_2"O}$

#### Explanation:

Start by assigning oxidation numbers to the atoms that take part in the reaction

${\stackrel{\textcolor{b l u e}{- 1}}{\text{Br")""^(-) + stackrel(color(blue)(+1))("H")_ 3 stackrel(color(blue)(+5))("As") stackrel(color(blue)(-2))("O")_ 4 -> stackrel(color(blue)(+5))("Br") stackrel(color(blue)(-2))("O")_ 3""^(-) + stackrel(color(blue)(+1))("H") stackrel(color(blue)(+3))("As") stackrel(color(blue)(-2))("O}}}_{2}$

Now, notice that the oxidation number of bromine is going from $\textcolor{b l u e}{- 1}$ on the reactants' side to $\textcolor{b l u e}{+ 5}$ on the products' side. This implies that bromine is being oxidized, i.e. its oxidation number is increasing.

Similarly, the oxidation number of astatine if going from $\textcolor{b l u e}{+ 5}$ on the reactants' side to $\textcolor{b l u e}{+ 3}$ on the products' side. This implies that astatine is being reduced, i.e. its oxidation number is decreasing.

The oxidation half-reaction looks like this

stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-)

Now, the bromine atoms are balanced. To balance the oxygen atoms, you can use water molecules on the side that needs oxygen. In this case, you need $3$ oxygen atoms on the reactants' side, so add $3$ water molecules to that side of the equation

$3 {\text{H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e}}^{-}$

To balance the hydrogen atoms, add protons, ${\text{H}}^{+}$, to the side that needs hydrogen. In this case, you have $6$ hydrogen atoms on the reactants' side, so add $6$ protons to the products' side

$3 {\text{H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-) + 6"H}}^{+}$

Notice that the charge is balanced because you have

$\left(1 -\right) = \left(1 -\right) + \left(6 -\right) + \left(6 +\right)$

The reduction half-reaction looks like this

${\text{H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O}}_{2}$

Once again, the astatine atoms are balanced, so focus on the oxygen atoms first. You will have

$\text{H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O"_ 2 + 2"H"_ 2"O}$

You now need $2$ protons on the reactants' side to balance the hydrogen atoms

$2 \text{H"^(+) + "H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O"_ 2 + 2"H"_ 2"O}$

Once again, notice that the charge is balanced because you have

$\left(2 +\right) + \left(2 -\right) = 0$

Now, in every redox reaction, the number of electrons lost in the oxidation half-reaction must be equal to the number of electrons gained in the reduction half-reaction.

This means that you must multiply the reduction half-reaction by $3$

{(color(white)(aaaaaaaa)3"H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-) + 6"H"^(+)), (2"H"^(+) + "H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 2"e"^(-) -> "H" stackrel(color(blue)(+3))("As") "O"_ 2 + 2"H"_ 2"O" " " | xx 3) :}#

to get

$\left\{\begin{matrix}\textcolor{w h i t e}{a a a a a a a a} 3 \text{H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 6"e"^(-) + 6"H"^(+) \\ 6"H"^(+) + 3"H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + 6"e"^(-) -> 3"H" stackrel(color(blue)(+3))("As") "O"_ 2 + 6"H"_ 2"O}\end{matrix}\right.$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

$3 \text{H"_ 2"O" + stackrel(color(blue)(-1))("Br")""^(-) + color(red)(cancel(color(black)(6"H"^(+)))) + 3"H"_ 3 stackrel(color(blue)(+5))("As") "O"_ 4 + color(red)(cancel(color(black)(6"e"^(-)))) -> stackrel(color(blue)(+5))("Br")"O"_ 3^(-) + 3"H" stackrel(color(blue)(+3))("As") "O"_ 2 + color(red)(cancel(color(black)(6"e"^(-)))) + color(red)(cancel(color(black)(6"H"^(+)))) + 6"H"_2"O}$

This will be equivalent to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{\text{Br"^(-) + 3"H"_3"AsO"_4 -> "BrO"_3^(-) + 3"HAsO"_2 + 3"H"_2"O}}}}$

Therefore, you can say that water appears in the balanced chemical equation as a product with a coefficient of $3$.

The bromide anion, ${\text{Br}}^{-}$ is being oxidized to the bromate anion, ${\text{BrO}}_{3}^{-}$, which implies that arsenic acid, ${\text{H"_3"AsO}}_{4}$, is acting as an oxidizing agent.

Consequently, you can say that arsenic acid is being reduced to arsonite, ${\text{HAsO}}_{2}$, which implies that the bromide anion is acting as a reducing agent.